Question

In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water.

Answer 1

Answer:

In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water

sodium carbonate + ethanoic acid †’ sodium ethanoate + carbon dioxide + water

Mass of sodium carbonate=5.3g (given)

Mass of ethanoic acid=6g(given)

Mass of carbon dioxide=2.2g(given)

Mass of water=0.9g(given)

Mass of sodium ethanoate=8.2g(given)

Now total mass before reaction=(5.3g+6g)=11.3g

Now total mass after reaction=(8.2g+2.2g+0.9g)=11.3g

Total mass before reaction=total mass after reaction

Therefore the given observation is in agreement with law of conservation of mass.

Answer 2

Answer:

According to the question,

Na

2

CO

3

+CH

3

COOH→CO

2

+H

2

O+CH

3

COONa

Mass of reactants =5.3+6=11.3 gm

Mass of products =2.2+0.9+8.2=11.3 gm

This shows, that during a chemical reaction, the mass of reactant = mass of the product.

Hence, it shows the law of conservation of mass.